MHB Combinations Problem - Confirming My Solution

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The discussion revolves around a combinations problem involving selecting groups from a committee of 12. The user calculates the total number of groups of four as 495 and proposes that the number of groups including either the chairperson or vice-chairperson, but not both, is 450. However, they are challenged by an alternative solution that suggests the correct answer is 240, based on choosing 3 additional members from the remaining 10 after selecting one of the two leaders. The user seeks confirmation of their calculations and understanding of the problem. The thread highlights the complexities of combinatorial logic in group selection scenarios.
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Hi everyone this is my first post. Please let me know if I am not following the proper etiquette.

I am helping out a friend with a problem. My answer to part (b) does not match the solution my friend has been given.

Four people are selected from a committee of 12 represent the committee at a particular function.
a) How many different groups of four are possible?
12C4 = 495
b) How many groups of four are possible of either the chairperson or the vice chairperson, but not both, must be in the group.
12C4 – 2C2 x 10C2 = 450
Alternatively
1C1 x 1C0 x 10C3 + 1C0 x 1C1 x 10C3 + 1C0 x 1C0 x 10C4
120 + 120 +210 = 450
I think this is correct, but the solution I have been given is 240?
 
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For both the chairperson and vice-chairperson, we find there are 10 remaining people from which to choose for the other 3 spots on the committee, so we find:

$$N=2{10 \choose 3}=240$$
 
MarkFL said:
For both the chairperson and vice-chairperson, we find there are 10 remaining people from which to choose for the other 3 spots on the committee, so we find:

$$N=2{10 \choose 3}=240$$

Many thanks! :cool:
 
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